Suppose $\mathcal C$ is a locally cartesian closed category (take simplicial sets for example), then for $B$ an object of $\mathcal C$, the slice category $\mathcal C/B$ is cartesian closed and thus has an exponential object $\textbf{Hom}_B(E_1,E_2)$.
If we then take $f:X\to B$, there is some kind of correspondence $$\text{Hom}_X(f^*E_1,f^*E_2)\cong \text{Hom}_B(X,\textbf{Hom}_B(E_1,E_2)).$$
In the context of simplicial sets, this would tell us what the simplices of $\textbf{Hom}_B(E_1,E_2)$ are characterized by pairs $(b:\Delta[n]\to B, u:b^* E_1\to b^* E_2)$.
The only way I could see this holds is by the identification $$\text{Hom}_B(X,\textbf{Hom}_B(E_1,E_2))\cong\text{Hom}_B(f^*E_1,E_2),$$
(because the product $X\times E_1$ in $\mathcal C/B$ is the pullback $f^*E_1$ as follows: $$ \require{AMScd} \begin{CD} f^*E_1@>>> E_1 \\ @VVV @VVV \\ X @>f>> B \end{CD} $$ ), and thus given a map $g:f^*E_1\to E_2$ coupled with the map $f^*E_1\to X$, we retrieve a unique map $u:f^*E_1\to f^*E_2$ by the universal property of pullbacks.
Conversely, given a map $u:f^*E_1\to f^*E_2$, we get a map $f^*E_1\to f^*E_2\to E_2$ by composing with the pullback projection map from $f^*E_2$.
I then have the following questions:
- Is this the right way to view the above correspondence?
- Is there a more simple way to see this that I am missing (ie. maybe using Yoneda in some way)?
You basically have it right. But naming objects of slices categories by their domain make you miss the obvious "simpler" way to see it. (It is simpler in the writting, not in the unpacking which is more or less what you wrote.)
Given $f: X \to B$ and $g_i: E_i \to B, (i=1,2)$, the hypothesis is that there is an object $\mathbf{hom}(g_1,g_2):\mathbf{hom}_B(E_1,E_2)\to B$ in the slice above $B$ together with a natural isomorphism: $$\mathcal C/B\,(f,\mathbf{hom}(g_1,g_2)) \simeq \mathcal C / B\,(f\times g_1,g_2) $$ As you remarked, for any $g$ with codomain $B$, $f\times g$ in $\mathcal C/B$ is defined as the pullback of $f$ and $g$ in $\mathcal C$. In particular it means that $f\times g \simeq f_!f^\ast (g)$ where $f_!$ denotes the postcomposition by $f$, which is left adjoint to $f^\ast: \mathcal C/B \to \mathcal C/X$. So you can conclude: $$\mathcal C/B\,(f,\mathbf{hom}(g_1,g_2)) \simeq \mathcal C / B\,(f_!f^\ast(g_1),g_2) \simeq \mathcal C/X\, (f^\ast(g_1),f^\ast(g_2))$$